A 1-qubit state is given by
$$ |\psi \rangle = a_{0} | 0 \rangle + a_{1} | 1 \rangle. $$
In order for the state to be physical, you must have
$$ |a_{0}|^{2} + |a_{1}|^{2} = 1. $$
You can represent the basis kets as column vectors:
$$ |0 \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |1 \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. $$
The three Pauli operators (\(X\), \(Y\), and \(Z\)) can be represented as matrices:
$$ X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$
It follows that
$$ X |0 \rangle = |1 \rangle, \qquad X |1 \rangle = |0 \rangle; $$
$$ Y |0 \rangle = i|1 \rangle, \qquad Y |1 \rangle = -i|0 \rangle; $$
$$ Z |0 \rangle = |0 \rangle, \qquad Z |1 \rangle = -|1 \rangle. $$
Since each of the Pauli matrices squares to the identity, the eigenvalues must be \(\pm 1\). The so-called "information/computational basis" (\(|0 \rangle\) and \(|1 \rangle\)) coincides with the eigen-basis of the \(Z\) operator:
$$ |Z+ \rangle = |0 \rangle, \qquad |Z- \rangle = |1 \rangle. $$
Thus, \(Z\) can be re-written as
$$ Z = |Z+ \rangle \langle Z+ | - |Z- \rangle \langle Z- | = |0 \rangle \langle 0 | - |1 \rangle \langle 1 |. $$
For \(X\), the normalized eigen-basis is
$$ |X+ \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle + |1 \rangle \right), \qquad |X- \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle - |1 \rangle \right). $$
The operator that takes a \(Z\) eigen-ket and returns an \(X\) eigen-ket is
$$ H = |X+ \rangle \langle 0 | + |X- \rangle \langle 1 |. $$
This is known as the Hadamard operator. Note that \(H\) is Hermitian.
For \(Y\), the normalized eigen-basis is
$$ |Y+ \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle + i|1 \rangle \right), \qquad |Y- \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle - i|1 \rangle \right). $$
The operator that takes a \(Z\) eigen-ket and returns a \(Y\) eigen-ket is
$$ G = |Y+ \rangle \langle 0 | + |Y- \rangle \langle 1 |. $$
Note that \(G\) is not Hermitian.
